01 · Which rate goes on top?
At a point, dx/dt = −4 and dy/dt = 6. Find dy/dx.
Hint
Divide the y-rate by the x-rate.
Worked solution
dy/dx = 6/(−4) = −3/2. Reversing the quotient would give dx/dy instead.
Understand · explore · practise
Parametric differentiation
Find dy/dx from parametric equations using dy/dt divided by dx/dt. Explore horizontal and vertical tangents, chain rules and points where both parameter derivatives vanish.
Before you startParametric coordinates; differentiation rules; radian trigonometry
01 / Divide the two parameter rates
A gradient compares changes in y with changes in x.
dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0
For x = f(t), y = g(t), first differentiate both coordinates with respect to the same parameter. Divide the y-rate by the x-rate. The answer is usually still expressed in t.
Changing the parameter can move a point left, right, up or down. The gradient is the ratio of the vertical rate to the horizontal rate, including their signs.
Two rates determine one gradientExplore
dx/dt = 2t; dy/dt = 3t² − 3.
t = 0.5; point = (1.25, −1.375); dx/dt = 1; dy/dt = −2.25; dy/dx = −2.25.
The rates have opposite signs, so the tangent gradient is negative.
The gold line follows the direction vector (dx/dt, dy/dt). When dx/dt is zero and dy/dt is nonzero, it is vertical. The parameter selects the point; it is not automatically the x-coordinate.
02 / Connect the formula to the chain rule
Both coordinates depend on the same changing parameter.
At a point where x can be inverted locallyWorked example
dy/dt = (dy/dx)(dx/dt)
Apply the chain rule to y as a function of x(t).
dy/dx = (dy/dt)/(dx/dt)
Divide only when the x-rate is nonzero.
A small change gives Δy/Δx = (Δy/Δt)/(Δx/Δt)
Taking the limit gives the same ratio when the denominator tends to a nonzero value.
The notation suggests cancellation of dt, but the justification is the chain rule or a limit. Do not cancel zero derivatives as if they were ordinary nonzero numbers.
Watch the two rates set the tangent direction
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Differentiate polynomial coordinates
Keep the selected parameter value until the final substitution.
x = t² + 1, y = t³ − 3tWorked example
dx/dt = 2t; dy/dt = 3t² − 3
Differentiate each coordinate separately.
dy/dx = (3t² − 3)/(2t), t ≠ 0
This is the gradient as a function of t.
At t = 2: point (5, 2), gradient 9/4
Substitute into both original coordinates and then the gradient formula.
02 · A negative parameter
For x = t² + 1, y = t³ − 3t, find the point and gradient at t = −2.
Hint
The same x-coordinate can occur at different parameter values.
Worked solution
The point is (5, −2). The gradient is 9/(−4) = −9/4.
03 · A linear x-coordinate
Find dy/dx if x = 3t − 2 and y = t² + 4t. Evaluate it at t = 1.
Hint
dx/dt is the constant 3.
Worked solution
dy/dx = (2t + 4)/3. At t = 1 the gradient is 2, and the point is (1, 5).
04 / Identify a horizontal tangent
The y-rate is zero while the x-rate is not.
Horizontal tangent: dy/dt = 0 and dx/dt ≠ 0
x = t² + 1, y = t³ − 3tWorked example
3t² − 3 = 0 gives t = ±1
Set the numerator rate to zero.
dx/dt = ±2, so both points are regular
The denominator is nonzero.
At t = −1: (2, 2); at t = 1: (2, −2)
Substitute separately to obtain both points.
04 · Find all horizontal tangents
For x = t³ + t, y = t² − 4t, find every point with a horizontal tangent.
Hint
dx/dt = 3t² + 1 is always positive.
Worked solution
dy/dt = 2t − 4 vanishes only at t = 2. The unique point is (10, −4); the tangent is horizontal there.
05 / Check a zero x-rate separately
A nonzero y-rate gives a vertical tangent.
x = t² + 1, y = t³ − 3t at t = 0Worked example
dx/dt = 0; dy/dt = −3
The direction vector is nonzero and vertical.
The point is (1, 0)
Use the original equations.
The tangent is x = 1; dy/dx has no finite value
Do not describe the undefined quotient as a finite gradient.
05 · Vertical on a circle
For x = 2 cos t, y = 2 sin t, with 0 ≤ t < 2π, find all points with a vertical tangent.
Hint
dx/dt = −2 sin t. At its zeros, check dy/dt = 2 cos t.
Worked solution
t = 0 and π give (2, 0) and (−2, 0). In both cases dy/dt is nonzero. The tangents are x = 2 and x = −2.
06 / Use radian trig derivatives
The parameter is an angle in radians.
x = 4 cos t, y = 3 sin tWorked example
dx/dt = −4 sin t; dy/dt = 3 cos t
Include each constant multiplier.
dy/dx = −3 cos t/(4 sin t)
This formula applies when sin t ≠ 0.
At t = π/4: point (2√2, 3√2/2), gradient −3/4
Use exact trig values.
06 · Horizontal on the ellipse
For the same ellipse, find the horizontal-tangent points over 0 ≤ t < 2π.
Hint
Set cos t = 0 and check sin t is nonzero.
Worked solution
t = π/2 and 3π/2 give (0, 3) and (0, −3). Both have gradient zero.
07 · Different inner angles
Find dy/dx for x = cos(2t), y = sin(3t).
Hint
Apply the chain rule separately to each coordinate.
Worked solution
dy/dx = −3 cos(3t)/[2 sin(2t)], wherever sin(2t) ≠ 0. Zeros of the denominator require separate analysis.
07 / Differentiate exponential and logarithmic coordinates
The parameter domain belongs to both coordinates.
x = eᵗ, y = t²Worked example
dx/dt = eᵗ; dy/dt = 2t
The x-rate is positive for every real t.
dy/dx = 2t/eᵗ
No real parameter is excluded by this denominator.
At t = 0: point (1, 0), gradient 0
A zero y-rate gives a regular horizontal tangent.
08 · A log coordinate
For x = ln t, y = t³, find dy/dx and its parameter domain.
Hint
The original logarithm requires t > 0.
Worked solution
dy/dx = 3t²/(1/t) = 3t³, for t > 0. Simplification does not extend the original parameter domain.
09 · Two changing exponentials
For x = e²ᵗ, y = eᵗ + t, find the gradient at t = 0.
Hint
The rates are 2e²ᵗ and eᵗ + 1.
Worked solution
dy/dx = (eᵗ + 1)/(2e²ᵗ). At t = 0 it equals 1; the point is (1, 1).
08 / Use product and quotient rules inside the coordinates
The outer ratio does not replace the rules inside each rate.
x = t², y = t ln t, with t > 0Worked example
dx/dt = 2t
Differentiate x.
dy/dt = ln t + 1
Use the product rule for y.
dy/dx = (ln t + 1)/(2t)
The denominator is nonzero on the original domain.
10 · A product coordinate
Find dy/dx for x = t + 1, y = t eᵗ.
Hint
The x-rate is 1; y needs the product rule.
Worked solution
dy/dx = eᵗ(1 + t), for all real t.
11 · A quotient coordinate
For x = t² + 1, y = t/(t + 1), find dy/dx where the quotient of rates is valid.
Hint
The y-rate is 1/(t + 1)².
Worked solution
dy/dx = 1/[2t(t + 1)²], for t ≠ −1, 0. The original curve excludes t = −1; at t = 0 it has a vertical tangent because dx/dt = 0 and dy/dt = 1.
09 / When both rates vanish, investigate the curve
The quotient 0/0 does not classify a tangent.
Compare two different parameterisations at t = 0Worked example
x = t³, y = t³: both derivatives are zero
The Cartesian curve is the straight line y = x, so its tangent slope is 1 even at the origin.
x = t³, y = t²: both derivatives are again zero
Here y = |x|^(2/3). The origin is a cusp with a vertical tangent.
Use elimination, a limit or a geometric argument
Two zero rates alone do not tell you whether the slope is finite or vertical.
12 · A removable quotient of rates
For x = t³, y = 2t³ + 1, investigate the tangent at t = 0.
Hint
Eliminate t before deciding what 0/0 means.
Worked solution
The curve is y = 2x + 1. It has slope 2 at (0, 1), even though both parameter derivatives vanish there. The rate quotient is 2 for t ≠ 0 and its limiting value agrees with the line.
10 / Cross-check by eliminating the parameter
Use a Cartesian equation when it is easy to obtain.
x = eᵗ, y = t²Worked example
t = ln x, with x > 0
Invert the x-coordinate.
y = (ln x)²
Substitute into y.
dy/dx = 2 ln x/x = 2t/eᵗ
The Cartesian and parametric derivatives agree.
13 · Two branches
Eliminate t from x = t² + 1, y = t. Check the parametric derivative on each branch.
Hint
x = y² + 1; for x > 1, y can be positive or negative.
Worked solution
Parametrically, dy/dx = 1/(2t) for t ≠ 0. The branches y = ±√(x − 1) have slopes ±1/[2√(x − 1)], agreeing with t = ±√(x − 1). At (1, 0) the tangent is vertical.
14 · A changed parameter speed
Compare x = t, y = t² with x = 2u, y = 4u² at the common point (2, 4).
Hint
Use t = 2 and u = 1. Each parameter gives different coordinate rates.
Worked solution
The first gives dy/dx = 2t = 4. The second gives (8u)/2 = 4. The parameter speed changes both rates, but their ratio gives the same tangent slope.
11 / Keep coordinates, rates and gradients distinct
Every division needs a nonzero denominator.
- Differentiate x and y with respect to the same parameter.
- Divide the y-rate by the x-rate.
- Substitute into the original equations to locate the point.
- For horizontal or vertical tangents, check both rates.
- If both rates vanish, analyse the curve separately.
- Preserve the original parameter domain and radian convention.
Section 1 of 11 · Divide the two parameter rates